Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{x^2 - 4x - 21}{5x - 35} \div \dfrac{10x + 30}{-6x + 6} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{x^2 - 4x - 21}{5x - 35} \times \dfrac{-6x + 6}{10x + 30} $ First factor the quadratic. $z = \dfrac{(x + 3)(x - 7)}{5x - 35} \times \dfrac{-6x + 6}{10x + 30} $ Then factor out any other terms. $z = \dfrac{(x + 3)(x - 7)}{5(x - 7)} \times \dfrac{-6(x - 1)}{10(x + 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (x + 3)(x - 7) \times -6(x - 1) } { 5(x - 7) \times 10(x + 3) } $ $z = \dfrac{ -6(x + 3)(x - 7)(x - 1)}{ 50(x - 7)(x + 3)} $ Notice that $(x - 7)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -6\cancel{(x + 3)}(x - 7)(x - 1)}{ 50(x - 7)\cancel{(x + 3)}} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $z = \dfrac{ -6\cancel{(x + 3)}\cancel{(x - 7)}(x - 1)}{ 50\cancel{(x - 7)}\cancel{(x + 3)}} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $z = \dfrac{-6(x - 1)}{50} $ $z = \dfrac{-3(x - 1)}{25} ; \space x \neq -3 ; \space x \neq 7 $